If PQR is an equilateral triangle inscribed in the auxiliary circle of the ellipse $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1(\mathrm{a}>\mathrm{b})$ and $P\text{'}Q\text{'}R\text{'}$ is corresponding triangle inscribed within the ellipse then centroid of the triangle $P\text{'}Q\text{'}R\text{'}$lies at

$$\begin{gathered} Let the ellipse be \frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b) \\ Equation of auxiliary circle is x^2+y^2=a^2 \end{gathered}$$

 Consider $\mathrm{P}\left(\mathrm{\theta }\right),\mathrm{Q}\left(\mathrm{\theta }+\frac{2\mathrm{\pi }}{3}\right)\text{ and }\mathrm{R}\left(\mathrm{\theta }+\frac{4\mathrm{\pi }}{3}\right)$ be three

points on auxiliary circle then corresponding vertices of triangle

whose vertices lie on ellipse are

$\begin{array}{l}{\mathrm{P}}^{\mathrm{\prime }}(\mathrm{acos} \mathrm{\theta },\mathrm{bsin} \mathrm{\theta }),\\ {\mathrm{Q}}^{\mathrm{\prime }}\left(\mathrm{acos} \left(\mathrm{\theta }+\frac{2\mathrm{\pi }}{3}\right),\mathrm{bsin} \left(\mathrm{\theta }+\frac{2\mathrm{\pi }}{3}\right)\right)\\ \mathrm{and} {\mathrm{R}}^{\mathrm{\prime }}\left(\mathrm{acos} \left(\mathrm{\theta }+\frac{4\mathrm{\pi }}{3}\right),\mathrm{bsin} \left(\mathrm{\theta }+\frac{4\mathrm{\pi }}{3}\right)\right)\end{array}$

Now centriod of triangle $$\begin{gathered} P\text{'}Q\text{'}R\text{'}=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})=(0,0) \\ Which is centre of ellipse \end{gathered}$$